MacLean_Jordan_Homework_1
Jordan MacLean
2024-01-16
https://rpubs.com/parker60/1138267
# List of packages
packages <- c("tidyverse", "fst", "modelsummary")
# Install packages if they aren't installed already
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)
# Load the packages
lapply(packages, library, character.only = TRUE)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors## [[1]]
## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
## [7] "tidyr" "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [13] "grDevices" "utils" "datasets" "methods" "base"
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## [1] "fst" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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## [[3]]
## [1] "modelsummary" "fst" "lubridate" "forcats" "stringr"
## [6] "dplyr" "purrr" "readr" "tidyr" "tibble"
## [11] "ggplot2" "tidyverse" "stats" "graphics" "grDevices"
## [16] "utils" "datasets" "methods" "base"# Doing so solves the issue of "Vector Memory" error, by telling R to clear its environment (memory)
rm(list=ls()); gc()## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1079765 57.7 2200462 117.6 NA 1405078 75.1
## Vcells 1861341 14.3 8388608 64.0 16384 2803286 21.4ess <- read_fst("All-ESS-Data.fst")#Task 1 #Provide code and answer. #Prompt and question: calculate the average for the variable ‘happy’ for the country of Norway. On average, based on the ESS data, who reports higher levels of happiness: Norway or Belgium? #Note: we already did it for Belgium. You just need to compare to Norway’s average, making sure to provide the code for both.
#First I will calculate the average for the varaible 'happy' in Norway, then I will do the same for Belgium
#This displays a table of all entered unique entries
unique(ess$cntry)## [1] "AT" "BE" "CH" "CZ" "DE" "DK" "ES" "FI" "FR" "GB" "GR" "HU" "IE" "IL" "IT"
## [16] "LU" "NL" "NO" "PL" "PT" "SE" "SI" "EE" "IS" "SK" "TR" "UA" "BG" "CY" "RU"
## [31] "HR" "LV" "RO" "LT" "AL" "XK" "ME" "RS" "MK"norway_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "NO") %>%
select(happy)norway_happy$y <- norway_happy$happy
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 15 29 59 163 238 730 817 2617 5235 3796 2344 12 10# need to remove 77 and 88 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 88 to NA
norway_happy$y[norway_happy$y %in% 77:88] <- NA
# checking again
table(norway_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 15 29 59 163 238 730 817 2617 5235 3796 2344#Now it is time to calculate the Mean
mean_y <- mean(norway_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.975005HERE IS THE ISSUE
# Now it is time to calculate the Median of 'y'
median_y <- median(norway_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8# Using tidyverse syntax principles, we can get both the mean and median at once.
norway_happy %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 7.975005 8### mean_y median_y
## 1 7.975005 8#Next, I will discover the Mode and Standard Deviation
mode_y <- norway_happy %>%
filter(!is.na(y)) %>%
count(y) %>%
arrange(desc(n)) %>%
slice(1) %>%
pull(y)
cat("\nMode of Y:", mode_y, "\n")##
## Mode of Y: 8##
## Mode of Y: 8sd_y <- sd(norway_happy$y, na.rm = TRUE)
cat("Standard Deviation of 'y':", sd_y, "\n")## Standard Deviation of 'y': 1.539186##
## Standard Deviation of Y: 1.539186# Let's compare if had not cleaned properly
norway_happy2 <- ess %>%
filter(cntry == "NO") %>%
select(happy)
sd_happy2 <- sd(norway_happy2$happy, na.rm = TRUE)
cat("Standard Deviation of 'happy not cleaned':", sd_happy2, "\n")## Standard Deviation of 'happy not cleaned': 3.146621## Standard Deviation of 'happy not cleaned': 3.146621 # Now I will summarize the happiness for Norway
summary(norway_happy)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 7.000 1st Qu.: 7.000
## Median : 8.000 Median : 8.000
## Mean : 8.076 Mean : 7.975
## 3rd Qu.: 9.000 3rd Qu.: 9.000
## Max. :88.000 Max. :10.000
## NA's :22#For the country of Norway, the average happiness score of the people surveyed is 7.975 on a scale of 0 to 10. This means that, on average, people tend to report their happiness level closer to the ‘happy’ end of the scale, as 10 represents “happy” and 0 represents “unhappy.”
#The standard deviation is 1.53, which tells us about the spread in happiness scores among respondents. A smaller standard deviation would mean that most people’s happiness scores are very close to the average, while a larger one indicates a wider range of scores. In this case, most people’s scores are within 1.52 points above or below our average score of 7.975. So, the majority of scores likely fall between roughly 6.44 (7.975 - 1.53) and 9.5 (7.975 + 1.53).
#With our ‘uncleaned’ variable we have a standard deviation of 3.14.
# Now I will compare and do the same for Belgium
belgium_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "BE") %>%
select(happy)belgium_happy <- ess %>% # note: if you work from belgium_data replace "ess" with belgium_data
filter(cntry == "BE") %>%
select(happy)belgium_happy$y <- belgium_happy$happy
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 50 27 104 194 234 830 999 3503 6521 3402 1565 3 16 3# need to remove 77, 88, 99 or else will alter results. See data portal for what they represent (e.g. DK, Refusal, etc.)
# Recode values 77 through 99 to NA
belgium_happy$y[belgium_happy$y %in% 77:99] <- NA
# checking again
table(belgium_happy$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 50 27 104 194 234 830 999 3503 6521 3402 1565# Now I will calculate the Mean and Median
mean_y <- mean(belgium_happy$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 7.737334## Mean of 'Y' is 7.737334# Now, let's determine the median of 'y'.
median_y <- median(belgium_happy$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n")## Median of 'y' is: 8## Median of 'Y' is 8# Using tidyverse syntax principles, we can get both the mean and median at once.
belgium_happy %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 7.737334 8### mean_y median_y
## 1 7.737334 8mode_y <- belgium_happy %>%
filter(!is.na(y)) %>%
count(y) %>%
arrange(desc(n)) %>%
slice(1) %>%
pull(y)
cat("\nMode of Y:", mode_y, "\n")##
## Mode of Y: 8## Mode of 'Y' is 8sd_y <- sd(belgium_happy$y, na.rm = TRUE)
cat("Standard Deviation of 'y':", sd_y, "\n")## Standard Deviation of 'y': 1.52045## Standard Deviation of 'Y' is 1.52045# Let's compare if had not cleaned properly
belgium_happy2 <- ess %>%
filter(cntry == "BE") %>%
select(happy)
sd_happy2 <- sd(belgium_happy2$happy, na.rm = TRUE)
cat("Standard Deviation of 'happy not cleaned':", sd_happy2, "\n")## Standard Deviation of 'happy not cleaned': 3.234511## Standard Deviation of 'happy not cleaned' is 3.234511#The average happiness score of the people surveyed is 7.737 on a scale of 0 to 10. This means that, on average, people tend to report their happiness level closer to the ‘happy’ end of the scale, as 10 represents “happy” and 0 represents “unhappy.”
#The standard deviation is 1.52, which tells us about the spread in happiness scores among respondents. A smaller standard deviation would mean that most people’s happiness scores are very close to the average, while a larger one indicates a wider range of scores. In this case, most people’s scores are within 1.52 points above or below our average score of 7.737. So, the majority of scores likely fall between roughly 6.2 (7.737 - 1.52) and 9.26 (7.737 + 1.52).
#With the ‘uncleaned’ variable we have a standard deviation of 3.23.
## Now, to see the results
summary(belgium_happy)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 7.000 1st Qu.: 7.000
## Median : 8.000 Median : 8.000
## Mean : 7.839 Mean : 7.737
## 3rd Qu.: 9.000 3rd Qu.: 9.000
## Max. :99.000 Max. :10.000
## NA's :22summary(norway_happy)## happy y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 7.000 1st Qu.: 7.000
## Median : 8.000 Median : 8.000
## Mean : 8.076 Mean : 7.975
## 3rd Qu.: 9.000 3rd Qu.: 9.000
## Max. :88.000 Max. :10.000
## NA's :22#Therefore, Norway has a higher average for the variable ‘happy’ with 7.975 and a standard deviation of ± 1.53, in comparison to the country of Belgium which possesses an average of 7.737 with a standard deviation of ± 1.52 .
Task 2
Provide code and answer.
Prompt and question: what is the most common category selected, for Irish respondents, for frequency of binge drinking? The variable of interest is: alcbnge.
ireland_alcbnge <- ess %>%
filter(cntry == "IE") %>%
select(alcbnge)
ireland_alcbnge$y <- ireland_alcbnge$alcbnge
table(ireland_alcbnge$y)##
## 1 2 3 4 5 6 7 8
## 65 650 346 417 239 641 26 6ireland_alcbnge$y[ireland_alcbnge$y %in% 6:8] <- NA# Now to calculate the mean, and the median
mean_y <- mean(ireland_alcbnge$y, na.rm = TRUE)
median_y <- median(ireland_alcbnge$y, na.rm = TRUE)
cat("Mean of 'y':", mean_y, "\n")## Mean of 'y': 3.066977cat("Median of 'y':", median_y, "\n")## Median of 'y': 3# Converting to categories to get mode as a category instead of a number
df <- ireland_alcbnge %>%
mutate(
y_category = case_when(
y == 1 ~ "Daily or almost daily",
y == 2 ~ "Weekly",
y == 3 ~ "Monthly",
y == 4 ~ "Less than monthly",
y == 5 ~ "Never",
TRUE ~ NA_character_
),
y_category = fct_relevel(factor(y_category), ### here you would put the categories in order you want them to appear or else it will appear alphabetically
"Daily or almost daily",
"Weekly",
"Monthly",
"Less than monthly",
"Never")
)
# To confirm the conversion:
table(df$y_category)##
## Daily or almost daily Weekly Monthly
## 65 650 346
## Less than monthly Never
## 417 239# Let's determine the mode of our newly created category:
get_mode <- function(v) {
tbl <- table(v)
mode_vals <- as.character(names(tbl)[tbl == max(tbl)])
return(mode_vals)
}
mode_values <- get_mode(df$y_category)
cat("Mode of y category:", paste(mode_values, collapse = ", "), "\n")## Mode of y category: Weekly#Mode of y category: WeeklyTherefore the most common category is Weekly, with 650 respondents.
Task 3
Provide code and answer.
Prompt and question: when you use the summary() function for the variable plnftr (about planning for future or taking every each day as it comes from 0-10) for both the countries of Portugal and Serbia, what do you notice? What stands out as different when you compare the two countries (note: look up the variable information on the ESS website to help with interpretation)? Explain while referring to the output generated.
# First I will compile the data for Portugal
portugal_plnftr <- ess %>%
filter(cntry == "PT") %>%
select(plnftr)portugal_plnftr$y <- portugal_plnftr$plnftr
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 88
## 114 184 313 356 264 481 262 382 345 166 370 40portugal_plnftr$y[portugal_plnftr$y %in% 87:88] <- NA # It's cool that the : is used as a through function
# checking again
table(portugal_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 114 184 313 356 264 481 262 382 345 166 370# now to get the mean
mean_y <- mean(portugal_plnftr$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 5.418289# Mean of 'y' is: 5.418289#now to get the median
median_y <- median(portugal_plnftr$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n") ## Median of 'y' is: 5#It is showcased that the median of 'y' is: 5#now to display both the mean and median together for Portugal
portugal_plnftr %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 5.418289 5summary(portugal_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 3.000 1st Qu.: 3.000
## Median : 5.000 Median : 5.000
## Mean : 6.426 Mean : 5.418
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :88.000 Max. :10.000
## NA's :14604 NA's :14644Now I will compile the data for Serbia, following the exact same steps
serbia_plnftr <- ess %>%
filter(cntry == "SE") %>%
select(plnftr)serbia_plnftr$y <- serbia_plnftr$plnftr
table(serbia_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88 99
## 151 155 398 517 301 503 270 405 374 145 237 2 7 1serbia_plnftr$y[serbia_plnftr$y %in% 77:99 ] <- NA
# checking again
table(serbia_plnftr$y)##
## 0 1 2 3 4 5 6 7 8 9 10
## 151 155 398 517 301 503 270 405 374 145 237# now to get the mean for Serbia
mean_y <- mean(serbia_plnftr$y, na.rm = TRUE)
cat("Mean of 'y' is:", mean_y, "\n")## Mean of 'y' is: 5.018229# Mean of 'y' is: 5.018229#now to get the median for Serbia
median_y <- median(serbia_plnftr$y, na.rm = TRUE)
cat("Median of 'y' is:", median_y, "\n") ## Median of 'y' is: 5#Median of 'y' is: 5#now to display both the mean and median together for Serbia
serbia_plnftr %>%
summarize(
mean_y = mean(y, na.rm = TRUE),
median_y = median(y, na.rm = TRUE)
) %>%
print()## mean_y median_y
## 1 5.018229 5#Once again it is evident that the median of 'y' is: 5#Now to compare the median of each
print(" Below represents the country Serbia")## [1] " Below represents the country Serbia"summary(serbia_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 3.000 1st Qu.: 3.000
## Median : 5.000 Median : 5.000
## Mean : 5.254 Mean : 5.018
## 3rd Qu.: 7.000 3rd Qu.: 7.000
## Max. :99.000 Max. :10.000
## NA's :14750 NA's :14760print("Below represents the country Portugal")## [1] "Below represents the country Portugal"summary(portugal_plnftr)## plnftr y
## Min. : 0.000 Min. : 0.000
## 1st Qu.: 3.000 1st Qu.: 3.000
## Median : 5.000 Median : 5.000
## Mean : 6.426 Mean : 5.418
## 3rd Qu.: 8.000 3rd Qu.: 8.000
## Max. :88.000 Max. :10.000
## NA's :14604 NA's :14644When it comes to using the summary() function for the variable plnftr about planning for the future or taking every each day as it comes from 0-10 for the countries of Portugal and Serbia, it is clear that while they both have the same median, the country of Portugal’s population (with a mean of 5.418) showcases more planning for the future than Serbia (5.018).
Task 4
#Provide code and answer.
#Prompt and question: using the variables stfdem and gndr, answer the following: on average, who is more dissastified with democracy in Italy, men or women? Explain while referring to the output generated.
#Info on variable here: https://ess.sikt.no/en/variable/query/stfdem/page/1
italy_data <- ess %>%
filter(cntry == "IT")
italy_data <- italy_data %>%
mutate(
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
TRUE ~ as.character(gndr)
),
stfdem = ifelse(stfdem %in% c(77, 88), NA, stfdem)
)mean_male_stfdem <- italy_data %>%
filter(gndr == "Male") %>%
summarize(mean_stfdem_men = mean(stfdem, na.rm = TRUE))
print(mean_male_stfdem)## mean_stfdem_men
## 1 4.782646# mean_male_stfdem
# 4.782646mean_female_stfdem <- italy_data %>%
filter(gndr == "Female") %>%
summarize(mean_stfdem_women = mean(stfdem, na.rm = TRUE))
print(mean_female_stfdem)## mean_stfdem_women
## 1 4.694652#mean_female_stfdem
# 4.694652# Compute average of stfdem by gender
means_by_gender <- italy_data %>%
group_by(gndr) %>%
summarize(stfdem = mean(stfdem, na.rm = TRUE))
print(means_by_gender)## # A tibble: 3 × 2
## gndr stfdem
## <chr> <dbl>
## 1 9 3.25
## 2 Female 4.69
## 3 Male 4.78Thus, using the variables stfdem and gndr, the population of men (4.782646 mean) are less disatisfied with democracy in Italy than women (4.694652 mean)
Task 5
#Provide code and answer.
#Prompt: Interpret the boxplot graph of stfedu and stfhlth that we generated already: according to ESS data, would we say that the median French person is more satisfied with the education system or health services? Explain.
#Change the boxplot graph: provide the code to change some of the key labels: (1) Change the title to: Boxplot of satisfaction with the state of education vs. health services; (2) Remove the x-axis label; (3) Change the y-axis label to: Satisfaction (0-10).
#Hint: copy the boxplot code above and just replace or cut what is asked.
france_data <- ess %>%
filter(cntry == "FR")# stfedu and stfhlth
france_data %>%
# Setting values to NA
mutate(stfedu = ifelse(stfedu %in% c(77, 88, 99), NA, stfedu),
stfhlth = ifelse(stfhlth %in% c(77, 88, 99), NA, stfhlth)) %>%
# Reshaping the data
select(stfedu, stfhlth) %>%
gather(variable, value, c(stfedu, stfhlth)) %>%
# Creating the boxplot
ggplot(aes(x = variable, y = value)) +
geom_boxplot() +
labs(y = "Satisfaction (0-10)", x = "", title = "Boxplot of satisfaction with the state of education vs. health services") +
theme_minimal()## Warning: Removed 364 rows containing non-finite values (`stat_boxplot()`).
The median French person is more satisfied with health services in comparison with the state of education, as it is evident that the median for education is 5.0, whereas the median of health services is roughly 7.3.